Chain rule for weak derivatives of $f(u)$ where $f'$ is not bounded but $u$ is?

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be $C^1$. Suppose $u$ has a weak derivative $u_x$. I want the chain rule $$\partial_x (f(u)) = f'(u)u_x$$ to hold. We know this holds if $f'$ is bounded. But I don't have that. But I do have $u$ is bounded almost everywhere, so $|f'(u)| \leq C$. Is this then enough for me to use the chain rule?

Answer 1

Definitely, it is!  Let $\Omega\subset\mathbb{R}^n$ be a bounded domain, $n\geqslant 1$, and $u\in W^{1,1}(\Omega)$. Function $\,u\,$ is additionally assumed to be essentially bounded on $\Omega$, so let $\,M=\|u\|_{L^{\infty}(\Omega)}\overset{\rm def}{=}\underset{x\in \Omega}{\rm ess\,sup\,}|u(x)|$.   Choose some cut-off function $\,\eta\in C_0^{\infty}(\mathbb{R})\,$ such that $$ \eta(t)= \begin{cases} 1,\quad |t|\leqslant M,\\ 0,\quad |t|\geqslant M+1, \end{cases} $$ and denote $\,g(t)\overset{\rm def}{=}\eta(t)\!\cdot\!f(t)$.  It is clear that function $\,g\in C^1(\mathbb{R})\,$ while $$ \sup_{t\in \mathbb{R}}\bigl(|g(t)|+|g'(t)|\bigr)\leqslant \max_{|t|\leqslant M+1}\bigl(|g(t)|+|g'(t)|\bigr), $$ i.e., function $g$ is bounded and uniformly Lipschitz on $\mathbb{R}$.   Notice that $\,f\bigl(u(x)\bigr)=g\bigl(u(x)\bigr)\,$ and $\,f'\bigl(u(x)\bigr)=g'\bigl(u(x)\bigr)\,$ a.e. in $\Omega$.   Hence, by virtue of the theorem 2.1.11 in Ziemer's textbook Weakly Differentiable Functions (see p. 48 therein) the chain rule $$ \partial_x f\bigl(u(x)\bigr)=\partial_x g\bigl(u(x)\bigr)= g'\bigl(u(x)\bigr)\!\cdot\!\partial_x u(x)= f'\bigl(u(x)\bigr)\!\cdot\!\partial_x u(x) $$ holds a.e. in $\Omega$.